15 picard iteration DAVID SEAL

PICARD ITERATIONDAVID SEAL The differential equation we’re interested in studying is = f t, y, yt0 = y0.. Many first order differential equations fall under this category and the followi

PICARD ITERATION

DAVID SEAL

The differential equation we’re interested in studying is

= f (t, y), y(t0) = y0 Many first order differential equations fall under this category and the following method is a new method for solving this differential equation The first idea is to transform the DE into an integral equation, and then apply a new method to the integral equation

We first do a change of variables to transform the initial conditions to the origin Explicitly, you can define w = y − y0 and x = t − t0 With a new f , the differential equation we’ll study is given by

= f (t, y), y(0) = 0

Note: it’s not necessary to do this substitution, but it makes life a lot easier if we do

Now, we integrate equation (2) from s = 0 to s = t to obtain

Z t s=0

y′ (s) ds =

Z t s=0

f(s, y(s)) ds

Applying the fundamental theorem of calculus, we have

Z t s=0

y′ (s) ds = y(t) − y(0) = y(t)

Hence we reduced the differential equation to an equivalent integral equation given by

Z t s=0

f(s, y(s)) ds

Even though this looks like it’s ‘solved’, it really isn’t because the function y is buried inside the integrand To solve this, we attempt to use the following algo-rithm, known as Picard Iteration:

(1) Choose an initial guess, y0(t) to equation (3)

(2) For n = 1, 2, 3, , set yn+1(t) =Rt

s=0f(s, yn(s)) ds Why does this make sense? If you take limits of both sides, and note that y(t) = limnyn+1= limnyn, then y(t) is a solution to the integral equation, and hence a solution to the differential equation The next question you should ask is under what hypotheses on f does this limit exist? It turns out that sufficient hypotheses are the f and fy be continuous at (0, 0) These are exactly the hypotheses given in your existence/uniqueness theorem 2

Date : September 24, 2009.

Note: If we stop this algorithm at a finite value of n, we expect yn(t) to be a very good approximate solution to the differential equation This makes this method of iteration an extremely powerful tool for solving differential equations!

For a concrete example, I’ll show you how to solve problem #3 from section

2 − 8

Use the method of picard iteration with an initial guess y0(t) = 0 to solve:

y′

= 2(y + 1), y(0) = 0

Note that the initial condition is at the origin, so we just apply the iteration to this differential equation

y1(t) =

Z t s=0

f(s, y0(s)) ds =

Z t s=0 2(y0(s) + 1) ds =

Z t s=0

2 ds = 2t

Hence, we have the first guess is y1(t) = 2t Next, we iterate once more to get y2:

y2(t) =

Z t

s=0

f(s, y1(s)) ds =

Z t s=0 2(y1(s) + 1) ds =

Z t s=0 2(2s + 1) ds =2

2 2!t

2 + 2t Hence, we have the second guess y2(t) = 2 2

2!t2+ 2t Iterate again to get y3:

y3(t) =

Z t

s=0

2(y2(s) + 1) ds =

Z t s=0

2 2 2 2!s

2+ 2s + 1



ds= (2t)

3 3! +

(2t)2 2! + 2t.

It looks like the pattern is

yn(t) =

n X k=1

(2t)k k!

and hence the exact solution is given by

y(t) = lim

n→∞yn(t) =

∞ X k=1

(2t)k k! =

n X k=0

(2t)k k! − 1 = e

2t

− 1

If you plug this into the differential equation, you’ll see we hit this one on the money To demonstrate this solution actually works, below is a graph of y5(t),

y (t) and y(t), the exact solution

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

−10

0

10

20

30

40

50

60

Approximate vs Exact Solution y

5(t) y

15(t) exact soln